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18.917 Topics in Algebraic Topology: The Sullivan Conjecture
Fall 2007
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Properties of T (Lecture 19)
Let V be a finite dimensional vector space over F2 . In this lecture, we will continue to establish some
of the basic properties of Lannes’ T-functor TV . More precisely, we will show that TV commutes with the
functor Φ and with the formation of tensor products.
To begin, we observe that for every unstable A-module M comes equipped with a canonical map
M → (TV M ) ⊗ H∗ (BV ).
This induces a map
ΦM → Φ(TV M ⊗ H∗ (BV )) � (ΦTV M ) ⊗ (Φ H∗ (BV )).
Composing with the Frobenius map Φ H∗ (BV ) → H∗ (BV ), we obtain a map
ΦM → (ΦTV M ) ⊗ H∗ M,
which is adjoint to a map
hM : TV ΦM → ΦTV M.
Proposition 1. For every unstable A-module M , the map hM : TV ΦM → ΦTV M is an isomorphism.
Proof. Choose a resolution
⊕β F (nβ ) → ⊕α F (nα ) → M → 0.
Since the functors TV and Φ both preserve cokernels and direct sums, we conclude that hM is an isomorphism
provided that the maps hF (n) are isomorphisms, for each n ≥ 0. We now work by induction on n, the case
n = 0 being obvious.
Recall that we have an exact sequence
0 → ΦF (n) → F (n) → ΣΩF (n) → 0.
Applying TV , we obtain another exact sequence
0 → TV ΦF (n) → TV F (n) → TV ΣΩF (n) → 0.
The functor TV commutes with Σ and Ω, so we can identify TV ΦF (n) with the kernel K of the unit map
TV F (n) → ΣΩTV F (n). On the other hand, we have an exact sequence
ΦTV F (n) → TV F (n) → ΣΩTV F (n) → 0,
which determines a surjective map g : ΦTV F (n) → K. The module TV F (n) is a direct sum of free unsta­
ble A-modules, and therefore reduced. It follows that g is also injective, and determines an isomorphism
ΦTV F (n) � TV ΦF (n). We now observe that this map is an inverse to hF (n) .
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We now discuss the behavior of TV with tensor products. Let M and N be unstable A-modules. We
have unit maps
M → TV M ⊗ H∗ (BV )
N → TV N ⊗ H∗ (BV ).
Tensoring these together and composing with the multiplication on H∗ (BV ), we get a map
M ⊗ N → TV M ⊗ TV N ⊗ H∗ (BV )
which has an adjoint
µM,N : TV (M ⊗ N ) → TV M ⊗ TV N.
Our goal is to prove the following:
Theorem 2. For every pair of unstable A-modules M and N , the map
µM,N : TV (M ⊗ N ) → TV M ⊗ TV N
is an isomorphism.
The proof proceeds in a series of steps. We begin with the following observation:
Remark 3. Let V = V0 ⊕ V1 . Then we have a canonical isomorphism
H∗ (BV ) � H∗ (BV0 ) ⊗ H∗ (BV1 ).
It follows that the functor M �→ M ⊗ H∗ (BV ) can be written as a composition of functors, given by tensor
product with H∗ (BV0 ) and H∗ (BV1 ) respectively. Passing to left adjoints, we get a canonical isomorphism
TV � TV0 ◦ TV1 .
The isomorphism of Remark 3 is compatible with the construction of the transformations µM,N . Conse­
quently, to prove Theorem 2, it will suffice to treat the case where V � F2 is one-dimensional.
Notation 4. If V = F2 , then we denote Lannes’ T-functor simply by T .
The following is a special case of Theorem 2:
Lemma 5. For every unstable A-module N , the canonical map
T (F (1) ⊗ N ) → T (F (1)) ⊗ T (N )
is an isomorphism.
Let us assume Lemma 5 for the moment, and use it to complete the proof of Theorem 2 in general.
Proof of Theorem 2. We wish to show that a canonical map
T (M ⊗ N ) → T (M ) ⊗ T (N )
is an isomorphism. As functors of M , both sides are compatible with the formation of cokernels and direct
sums. We may therefore argue as in the proof of Proposition 1 to reduce to the case where M � F (m) is
a free module. Recall that F (m) is canonically isomorphic to Σm -invariants in the tensor product F (1)⊗m .
Since the functor T is exact, it commutes with the formation of fixed points. It will therefore suffice to prove
the result in the case M = F (1)⊗m . We have a commutative diagram
µM,N
� T (F (1)⊗m ) ⊗ T (N )
T (F (1)⊗m ⊗ N )
���
��� µ�
���
µ�
���
���
�
T (F (1))⊗m ⊗ T (N ).
It follows from repeated application of Lemma 5 that the maps µ� and µ�� are isomorphisms, so that µM,N
is an isomorphism as well.
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Proof of Lemma 5. We wish to show that the canonical map
µF (1),N : T (F (1) ⊗ N ) → T (F (1)) ⊗ T (N )
is an isomorphism. As functors of N , both sides preserve direct sums and cokernels. We may therefore
assume that N � F (n) is a free unstable A-module. We proceed by induction on n. We need to prove three
things:
(a) The map µF (1),N is an isomorphism in every positive degree k. To prove this, we observe that N is
reduced, so we have a map of exact sequences
0 → T (F (1) ⊗ ΦN )
µF (1),ΦN
�
0 → T (F (1)) ⊗ T (Φ(N ))
� T (F (1) ⊗ N )
�
µF (1),N
� T (F (1)) ⊗ T (N )
� T (F (1) ⊗ ΣF (n − 1))
�
�0
µF (1),ΣF (n−1)
� T (F (1)) ⊗ T (ΣF (n − 1))
� 0.
Since T commutes with suspension, the inductive hypothesis guarantees that µF (1),ΣF (n−1) is an iso­
morphism. Consequently, to show that µF (1),N is an isomorphism in degree k, it will suffice to show
that µF (1),ΦN is an isomorphism in degree k. We have a second map of exact sequences
0 → T (ΦF (1) ⊗ ΦN )
µΦF (1),ΦN
�
0 → T (ΦF (1)) ⊗ T (Φ(N ))
� T (F (1) ⊗ ΦN )
µF (1),ΦN
�
� T (F (1)) ⊗ T (ΦN )
� T (ΣF (0) ⊗ ΦN )
�
�0
µΣF (0),ΦN
� T (ΣF (0)) ⊗ T (ΦN )
� 0.
Since T commutes with Σ, the map µΣF (0),ΦF (n) is an isomorphism. Consequently, to prove that
µF (1),N is an isomorphism in degree k, it will suffice to show that µΦF (1),ΦN is an isomorphism in
degree k. Since T commutes with Φ, this is equivalent to the assertion that µF (1),N is an isomorphism
in degree k2 , which follows from the inductive hypothesis.
(b) The map µF (1),N is surjective in degree 0. For each p ≥ 0, the vector space (T F (p))0 is dual to
HomA (T F (p), J(0)) � HomA (F (p), H∗ (BF2 )) � Hp (BF2 ).
In particular, it is a one-dimensional vector space over F2 , generated by tp ∈ H∗ (BF2 ) � F2 [t]. It
follows that T (F (1)) ⊗ T (F (n)) is also one-dimensional in degree 0. Moreover, in degree zero the map
µF (1),N is dual to the composition
F2 � HomA (T (F (1)) ⊗ T (N ), J(0)) → HomA (T (F (1) ⊗ N ), J(0)) � HomA (F (1) ⊗ N, H∗ (BF2 )).
We wish to show that this map is injective. For this, it suffices to observe that the image of the
nontrivial element of F2 is a homomorphism F (1) ⊗ N → H∗ (BF2 ) given by multiplying the nontrivial
maps F (1) → H∗ (BF2 ) and N → H∗ (BF2 ), and that this map is nontrivial in degree n + 1.
(c) The map µF (1),N is injective in degree zero. Given (b) and the observation that T (F (1)) ⊗ T (N ) is
one-dimensional in degree 0, it will suffice to show that the dimension of T (F (1) ⊗ N )0 is at most 1.
We will prove the following more general assertion:
(∗p ) The dimension of T (Φp F (1) ⊗ F (n))0 is at most 1.
For p large, we will invoke the following lemma:
Lemma 6. Fix an integer n. Then for p � 0, the tensor product Φp F (1) ⊗ F (n) is generated by a
single element.
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Assuming Lemma 6, we deduce that for p � 0 we have a surjection F (m) → Φp F (1) ⊗ F (n). This
induces a surjection
F (m) ⊕ F (m − 1) ⊕ . . . ⊕ F (0) � T F (m) → T (Φp F (1) ⊗ F (n)).
Since the left hand side has dimension 1 in degree 0, assertion (∗p ) follows.
To prove (∗p ) in general, we use descending induction on p. We have an exact sequence
p
0 → Φp+1 F (1) ⊗ F (n) → Φp F (1) ⊗ F (n) → Σ2 F (n) → 0
Since T is an exact functor which commutes with Σ, this reduces to an isomorphism T (Φp+1 F (1) ⊗
F (n))0 � T (Φp F (1) ⊗ F (n))0 , so that (∗p+1 ) implies (∗p ) as desired.
We will give the proof of Lemma 6 in the next lecture.
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